One direction follows by restriction. Conversely, suppose $|\cdot|$ is a <Non-Archimedean absolute value>. Then $|m|_L=|m|\leq1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality give
$$
|x+y|_L^n
\leq\sum_{j=0}^n\binom nj|x|_L^j|y|_L^{n-j}
\leq(n+1)\max(|x|_L,|y|_L)^n.
$$
Taking $n$th roots and letting $n\to\infty$ yields the <ultrametric inequality> for $|\cdot|_L$. Thus an <extension of an absolute value> is non-Archimedean exactly when its restriction is.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 4 / a / i / Solution

For a <field extension> $L/K$, an extension of an absolute value $|\cdot|$ on $K$ is an absolute value on $L$ whose restriction to $K$ is $|\cdot|$.

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 Incoming links: Extension of an absolute value